Question

If $a+\frac{1}{a}=\sqrt{3}$, then value of $a^3+\frac{1}{a^3}$ is:

Answer 1

We know, $(a+b{)}^3=a^3+b^3+3ab(a+b)$

$⟹$ $a^3+b^3=(a+b{)}^3-3ab(a+b)$.

Then,

$a^3+\frac{1}{a^3}={(a+\frac{1}{a})}^3-3(a+\frac{1}{a})$

$=(\sqrt{3}{)}^3-3\sqrt{3}=0$

$=3\sqrt{3}-3\sqrt{3}=0$.